Problem: Let $R$ be the region in the first quadrant enclosed by the $x$ -axis, the $y$ -axis, the line $y=2$, and the curve $y=\sqrt{9-x^2}$. $y$ $x$ ${y=\sqrt{9-x^2}}$ $ R$ $ 0$ $ 2$ A solid is generated by rotating $R$ about the $y$ -axis. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Explanation: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\sqrt{9-x^2}}$ Notice the slices are horizontal, because we are rotating $R$ about the $y$ -axis. Each slice is a cylinder. Let the thickness of each slice be $dy$ and let the radius of the base, as a function of $y$, be $r(y)$. Then, the volume of each slice is $\pi [r(y)]^2\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(y)]^2\,dy$ This is called the disc method. What we now need is to figure out the expression of $r(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\sqrt{9-x^2}}$ $ 0$ $ 2$ $r$ The radius is equal to the distance between the curve $y=\sqrt{9-x^2}$ and the $y$ -axis. To find it, we need to solve the equation for $x$ : $x=\sqrt{9-y^2}$ So, for any $y$ -value, $r(y)=\sqrt{9-y^2}}$. Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(y)}]^2 \\\\ &=\pi\left(\sqrt{9-y^2}}\right)^2 \\\\ &=\pi(9-y^2) \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=2$. So the interval of integration is $[0,2]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^2 \left[\pi\left(9-y^2\right)\right]dy \\\\ &=\pi \int^2_0 \left(9-y^2\right)\, dy \end{aligned}$ Let's evaluate the integral. $\pi \int^2_0 \left(9-y^2\right)\, dy=\dfrac{46\pi}{3}$ In conclusion, the volume of the solid is $\dfrac{46\pi}{3}$.